Can you help me please

) e number of viewers ordering a particular pay-per-view program is normally distributed. Past history shows that 33.00% of the

lubasha [3.4K]

Answer:

The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Past history shows that 33.00% of the time fewer than 20,000 people order the program

This means that X = 20000 has a pvalue of 0.33. So when X = 20000, Z = -0.44.

$Z = \frac{X - \mu}{\sigma}$

$-0.44 = \frac{20000 - \mu}{\sigma}$

$20000 - \mu = -0.44\sigma$

$\mu = 20000 + 0.44\sigma$

Only ten percent of the time do more than 28,000 people order the program.

This means that X = 28000 has a pvalue of 1-0.1 = 0.9. So when X = 28000, Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{28000 - \mu}{\sigma}$

$28000 - \mu = 1.28\sigma$

$\mu = 28000 - 1.28\sigma$

Also

$\mu = 20000 + 0.44\sigma$

So

$20000 + 0.44\sigma = 28000 - 1.28\sigma$

$1.72\sigma = 8000$

$\sigma = \frac{8000}{1.72}$

$\sigma = 4651.16$

$\mu = 20000 + 0.44\sigma = 20000 + 0.44*4651.16 = 22046.5$

The mean of the number of people ordering the program is 22,046.5 and the standard deviation is 4,651.16.

6 0

3 years ago

In the input/output rule: y = -2x + 8 What

Doss [256]

Answer:

"input" is x and "output" is y

y= -2(3) +8

y= -6 +8

y= 2

your answer would be 2

3 0

3 years ago

A bag of grass seeds wieghed 1/8 of a gram. that was enough to 1/6 of a lawn with seed. how many bags would it take to completly

bogdanovich [222]

According to the text, 1 bag of grass seed = 1/6 of the lawn. So if this is right, then it will take 6 bags of grass seeds to cover 1 lawn.

8 0

3 years ago

On a particular day, 112 of 280 passengers on a particular DTW-LAX flight used the e-ticket check-in kiosk to obtain boarding pa

Sonja [21]

Answer:

0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

Step-by-step explanation:

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Population of 280, which means that $N = 280$

112 use e-ticket check-in kiosk, which means that $k = 112$

Sample of eight passengers means that $n = 8$

Find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

This is P(X = 4). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,280,8,112) = \frac{C_{112,4}*C_{168,4}}{C_{280,4}} = 0.2348$

0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

3 0

3 years ago

Point S is on line segment \overline{RT}

BartSMP [9]

Answer:1

Step-by-step explanation:

5 0

4 years ago

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